Eyepieces and magnification

Builds on: Sections 5.1 and 5.2
Reading time: 35 min
Exercises: 5
Olympiad problems: 1

Through the telescope of IOAA 2026 with a 25 mm eyepiece, Jupiter, about 45″ across, looks as large as an object 22.5′ across seen with the naked eye, a little smaller than the full Moon looks without a telescope. With a 10 mm eyepiece it looks 2.5 times larger again, but dimmer, and on a night of steady air it shows more detail. In this section we see how the eyepiece magnifies, why every telescope has a lowest and a highest useful magnification, how much sky an eyepiece shows, and how a small concave lens called a Barlow lens raises the magnification without making the telescope much longer.

5.3.1How an eyepiece magnifies

Before you readMake a guess first

The same telescope is used with an eyepiece of focal length 25 mm and then with one of 10 mm. Which gives the higher magnification?

The objective forms a small image of the sky in its focal plane, as Section 5.1 showed. An is a small lens, or a group of lenses, placed so that its own focus falls on that image. Light from each point of the image then leaves the eyepiece as a parallel beam, which a relaxed eye focuses without strain. A star an angle $θ$ from the centre of the field forms an image $F θ$ from the centre of the focal plane, and the eyepiece sends its light out at an angle $F θ / f_{e}$ . Every angle on the sky is therefore multiplied by $F / f_{e}$ , the .

ResultMagnification and length of a telescope(5.5)

\begin{aligned} M & = \frac{F}{f_{e}} \\ L & = F + f_{e} \end{aligned}

$M$	the angular magnification, the ratio of the angle seen through the telescope to the angle on the sky
$F, f_{e}$	the focal lengths of the objective and of the eyepiece
$L$	the distance between a refractor's objective and its eyepiece when both foci meet, which is the length of the telescope

The image is upside down, which does not matter for astronomy. Changing the eyepiece is the usual way to change the magnification.

Quick checkThe 25 mm eyepiece

What magnification does a 25 mm eyepiece give on the telescope of IOAA 2026, focal length 750 mm?

times

5.3.2The exit pupil

All the light collected by an objective of diameter $D$ leaves the eyepiece as a beam $D / M$ wide, called the exit pupil. It is the small bright disc that we see floating behind the eyepiece when the telescope points at the daytime sky. Since $M = F / f_{e}$ , the exit pupil is also $f_{e} / N$ , where $N$ is the focal ratio.

ResultExit pupil(5.6)

d = \frac{D}{M} = \frac{f_{e}}{N}

$d$	the width of the beam leaving the eyepiece, the exit pupil
$D, N$	the diameter and the focal ratio of the objective

If the exit pupil is wider than the pupil of the eye, about 7 mm in the dark, some of the light misses the eye. The lowest useful magnification is therefore about

D / 7 mm

Figure 5.2The beam leaving an eyepieceinteractive

Eyepiece f_e

Magnification
Exit pupil
True field with a 52° eyepiece

The figure shows the telescope of IOAA 2026, 150 mm across with a focal length of 750 mm, with the part behind the dotted line, after the focus, drawn about seven times larger than the rest so that the beam leaving the eyepiece can be seen. Light from a star on the axis meets at the focus P of the mirror, spreads again, and leaves the eyepiece E as a parallel beam, drawn heavy, whose width is the exit pupil. The dashed bracket on the right is the 7 mm pupil of a dark-adapted eye. The slider sets the eyepiece's focal length. A gives an exit pupil of 8 mm, wider than the eye can take in, a gives 5 mm, and a gives a beam only 1.2 mm wide at a magnification of 125.

There is also a highest useful magnification. Observers have long used a rule that it is about twice the diameter of the objective in millimetres. At that magnification the finest detail the telescope can resolve (Section 5.4) is enlarged to a few arcminutes, which is 4.6′ for a 150 mm telescope at 300 times, several times the 1′ the eye can see and so easy to see. Beyond it the image only grows larger and dimmer, with the blur of each star enlarged along with it. On most nights the moving air sets a lower limit still, as Section 5.4 explains.

5.3.3Fields of view

An eyepiece shows a circle of the focal plane, and the angle across that circle as it appears to the eye is the eyepiece's apparent field, usually 40° to 80°. The true field is the angle of sky that the circle covers, and since every angle is multiplied by $M$ , the true field is the apparent field divided by the magnification. The telescope of IOAA 2026 with a 25 mm eyepiece of apparent field 52° shows $52 ° / 30 = 1.73 °$ of sky, enough to hold the whole Pleiades cluster.

When the apparent field is not known, the true field can be measured by timing a star as it drifts across the field with the telescope's drive switched off. A star at declination $δ$ moves ${15.04}^{″} \cos δ$ each second (Section 5.2), so a star that takes $t$ seconds to cross the middle of the field from edge to edge gives a true field of ${15.04}^{″} \times t \cos δ$ . This drift method is a common task in the telescope round.

Example 5.5Timing Vega across the field

With the drive switched off, Vega, at declination +38.8°, takes 4 min 30 s to drift across the middle of the field of an eyepiece. What is the true field?

Solution

Vega moves ${15.04}^{″} \times \cos 38.8 ° = {11.72}^{″}$ each second, so in 270 s it moves

270 \times {11.72}^{″} = 3164^{″} = {52.7}^{'} .

AnswerThe true field is 52.7′, a little less than a degree.

5.3.4The Barlow lens

A Barlow lens is a concave lens placed in the cone of light a short way in front of the focus. It makes the cone less steep, so the light meets further back and the image there is larger, as if the objective had a longer focal length. A Barlow lens that doubles the image doubles the magnification of every eyepiece, while the telescope grows only a little longer. The problem below, from IOAA 2016, finds where the lens must go.

Common errorsSection 5.3

A higher magnification does not always show more. Above about twice the aperture in millimetres, or above what the air allows that night, the image only grows dimmer and more blurred.
The apparent field belongs to the eyepiece and the true field to the sky. The true field is the apparent field divided by the magnification.
In the drift method the star must cross the middle of the field. A star that crosses near the edge cuts a shorter chord and gives too small a field.

ExercisesSection 5.35 questions

5.3.1
What magnification does a 10 mm eyepiece give on a telescope of focal length 750 mm?
times
5.3.2
What is the exit pupil of a 150 mm telescope at a magnification of 75?
mm
5.3.3
What is the lowest useful magnification of a telescope 200 mm across, for an eye whose pupil opens to 7 mm?
times
5.3.4
An eyepiece with an apparent field of 68° gives a magnification of 75. What is its true field, in arcminutes?
arcminutes
5.3.5
A star on the celestial equator takes 2 min 40 s to drift across the middle of an eyepiece's field. What is the true field, in arcminutes?
arcminutes

Olympiad problemsPast IOAA problems, restated1 problem

2016A telescope with a Barlow lens20 points

Adapted from IOAA 2016 (Bhubaneswar), theory problem 7. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

A refracting telescope of focal ratio f/5 has an objective of focal length 100 cm and an eyepiece of focal length 1 cm.

(1) What is its angular magnification $m_{0}$ , and what is its length $L_{0}$ , the distance from objective to eyepiece? (4 points)

To double the magnification, a Barlow lens of focal length −1 cm is placed between the objective and the focus of the objective.

(2) How far in front of the objective's focus must the Barlow lens be placed? (6 points)

(3) By how much does the telescope become longer? (4 points)

(4) The eyepiece and the Barlow lens are removed, and a camera with pixels 10 µm wide is placed at the objective's focus. How many pixels apart are the images of two stars 20″ apart? (6 points)

I solved it without the solution

AnswerThe telescope magnifies 100 times and is 101 cm long. The Barlow lens goes 0.5 cm in front of the objective's focus, the telescope becomes 0.5 cm longer, and the two stars are about 10 pixels apart on the camera.